Asked by Dan
The function f is defined by f(x)=4x^3+2ax^2+x+b, where a and b are constants.
i) Given that x^2-3x-4 is a factor of f(x), find the value of a and of b.
ii) Hence, factorise f(x) completely.
i) Given that x^2-3x-4 is a factor of f(x), find the value of a and of b.
ii) Hence, factorise f(x) completely.
Answers
oobleck
x^2 - 3x - 4 = (x-4)(x+1), so both (x-4) and (x+1) are factors of f(x)
so, f(4) = f(-1) = 0
That means we have
256+32a+4+b = 0
-4+2a-1+b = 0
so a = -53/6 and b = 68/3
f(x) = 4x^3 - 53/3 x^2 + x + 68/3 = (x-4)(x+1)(4x - 17/3)
so, f(4) = f(-1) = 0
That means we have
256+32a+4+b = 0
-4+2a-1+b = 0
so a = -53/6 and b = 68/3
f(x) = 4x^3 - 53/3 x^2 + x + 68/3 = (x-4)(x+1)(4x - 17/3)